Problem: Gandhijee is interested in building human tree. He defined a human node as follows :
Person_Id = English alphabet {a...z} .          
Person_Chain_Definition = Person_Id ( Person_Chain_Definition   Person_Chain_Definition )

For example : 
a( b(..) c( d( . e(..) ) f( . . ) ) ) refers to a human tree having a as the root human, 
whose chain links are are b and c, (spaces here are for clarity, input file has no spaces).   

Note : After every Person_Id there is necessarily an opening parenthesis ( that starts the definition of the sub-tree under the person. For empty(NULL) human nodes, there is a '.' character. 
Now, Gandhijee figure out the tree as follows : 
Step 1 : He writes root human root id in column 0. 
Step 2 : Then he writes left link's person id in 1 column to the left and right link's person id to 1 column to the right. 
He continues writing the subtree, in a way that, for an id written in column C, it's left link's id is written in column C-1 and the right link's id is written in column C+1.   

Now, Gandhijee wants to know the id's of people standing in a particular column C in lexicographically sorted order.
You are provided with column number C and the Tree definition string S. Gandhijee is very busy and hence he wants you to help him.   

Input : 
First line contains a number T, the number of test cases. 
Each test case consists of a single line containing C, the column number for which you need to print the result, followed by a space followed by definition of the tree in the format given above, without any spaces.

Output : 
For each test case, on a new line print the required id's in lexicographically sorted order, without spaces.  
If there are no id's in column C, print "Common Gandhijee!" (quotes for clarity).   

Constraints : 
1 ≤ T ≤ 100  
1 ≤ Len(S) ≤ 10000 
-10000 ≤ C ≤ 10000

Image for given testcase.      SAMPLE INPUT
2
-1 a(c(f(.h(..))b(g(..).))e(.d(..)))
3 b(c(..)a(..))

SAMPLE OUTPUT
cgh
Common Gandhijee!

Explanation

For the 1st sample case, from the diagram clearly c, g, h lies in column -1.

Solution (in PYTHON):

Solution:
#!/usr/bin/python
#gandhi tree problem
import copy
def convert(key, mystr):
	global column,temp,Fcolumn
	L = list(mystr)
	#check if binary or single
	open = 0
	close = 0
	LFindex = 0
	LLindex = 0
	RFindex = 0
	RLindex = 0
	#stores the a of a()
	column[key] = L.pop(0)
	#removes '(' and ')'
	L.pop(0)
	L.pop(len(L)-1)
	if len(L) == 0:
		return
	mystr = mystr[2:] + mystr[:len(mystr)]
	for i, ch in enumerate(L):
		if L[i] == '(':
			open = open + 1
			if open == 1:
				LFindex = i-1
			continue	
		if L[i] == ')':
			close = close - 1
			if open == abs(close):
				LLindex = i
				break;

	if not i == len(L) - 1: 
		# it means there is  right tree
		RFindex = LLindex + 1
		RLindex = len(L) - 1
	
	#here key receives CR CL CRL means Center Right Left like
	lkey = key + "L"
	lstr = mystr[LFindex:LLindex+1]
	column[lkey] = lstr
	rkey = key + "R"
	rstr = mystr[RFindex:RLindex+1]
	column[rkey] = rstr
treeDisp = []
indDisp = []
map_column = {"C":0, "L":-1, "R":1}
T = int( raw_input() )
for count in range(T):
	column = {}
	Fcolumn = {}
	temp = {}
	Gandhi_Tree = {}
	colDisp, line = raw_input().split(" ")
	colDisp = int(colDisp)
	indDisp.append(colDisp)
	line = line.replace("..","")
	line = line.replace(".","@()")
	convert("C",line)
	while len(column):
		keyall = column.keys()
		if not len(keyall):
			convert(keyall[0],column[keyall[0]])
		else:
			if len(column[keyall[0]]) == 1:
				Fcolumn[keyall[0]] = column.pop(keyall[0])
				continue
			convert(keyall[0],column[keyall[0]])
			Fcolumn[keyall[0]] = column.pop(keyall[0])
	column.clear()
	temp.clear()
	for key in Fcolumn:
		newkey = 0
		for ch in key:
			newkey = newkey + map_column[ch]
		if Gandhi_Tree.has_key(newkey):
			Gandhi_Tree[newkey] += Fcolumn[key].replace('@','')
		else:
			Gandhi_Tree[newkey] = Fcolumn[key].replace('@','')
	treeDisp.append(Gandhi_Tree)
for ind, tree in zip(indDisp, treeDisp):
	mystr = ""
	try:
		sorted_output = list(tree[ind])
		sorted_output.sort()
		for ch in sorted_output:
			mystr = mystr + ch
	except KeyError:
		mystr = "Common Gandhijee!"
	if mystr == "":
		print "Common Gandhijee!"
	else:
		print mystr